Solve the system of linear equations using matrices. x+y+z=3 2x+3y+2z=7 3x-4y+z=4

Kaycee Roche 2021-02-13 Answered
Solve the system of linear equations using matrices.
x+y+z=3
2x+3y+2z=7
3x-4y+z=4
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Expert Answer

insonsipthinye
Answered 2021-02-14 Author has 83 answers
Step 1
Given equation
x+y+z=3
2x+3y+2z=7
3x-4y+z=4
in matrix form
A=[111232341]
B=[374]
X=[xyz]
we know that AX=B
So, X=A1B
Step 2
Now,
X=A1B
We have to find A1
Adjoin the identity matrix onto the right of the original matrix, so that you have A on the left side and the identity matrix on the right side. [111232341]=[100010001]
now we will apply the elementary row method R22R1R2 (multiply 1 row by 2 and subtract it from 2 row), R33R1R3 (multiply 1 row by 3 and subtract it from 3 row)
[111221321221331431131]=[100021120020031030130][111010072]=[100210301]
R11R2R1 (multiply 2 row by 1 and subtract it from 1 row), R3+7R2R3 (multiply 2 row by 7 and add it to 3 row)
[101010002]=[3102101771]
R32R3 (divide the 3 row by -2)
[101010001]=[3102108.53.50.5]
R11R3R1 (multiply 3 row by 1 and subtract it from 1 row)
[100010001]=[5.52.50.52108.53.50.5]
A1=[5.52.50.52108.53.50.5]
Step 3
Now,
X=A1B
Jeffrey Jordon
Answered 2022-01-27 Author has 2047 answers

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